Combined Heat and Power
Introduction
Thomas Newcomen’s steam engine in 1712 launched the Industrial Revolution by
converting heat to power. Sadi Carnot improved the understanding of these new heat
engines by postulating that heat engines require more heat input than work output, and
thus some waste heat must always be rejected.
Newcomen’s steam pump
Carnot’s heat engine
Building on these fundamental relationships between heat and power, modern
combined heat and power (CHP) technologies:
 utilize the heat produced during power production to heat buildings or
manufacturing processes, or
 utilize heat produced during manufacturing processes to produce useful power.
In both of the above cases, utilizing heat that would otherwise be wasted increases the
overall efficiency of energy utilization. This increased energy efficiency drives
substantial interest in increasing the number of CHP applications. For example, Ohio
provides an interesting snap shot of CHP potential. As of 2012, total electrical power
output of installed CHP applications in Ohio is 766.6 MW. This represents 2.3% of all
electrical power generation. Nationwide, the average statewide fraction of CHP
electrical power to total electrical power is 8%.
1
Source: Midwest Clean Energy Application Center www/midwestcleanenergy.org
DOE estimates the technical potential of CHP to be 9,800 MW. Thus, in terms of
technical potential, Ohio is rated as 5th. However, in terms of implementation, Ohio is
rated as 43rd.
2
Candidates for Combined Heat and Power
Cogeneration equipment is expensive and is seldom cost-effective for plants with small
electrical and/or thermal loads. In addition, it is seldom economical for a plant to
generate its own electricity unless the plant can also use the heat rejected from the
electrical generation process. Moreover, because of the relatively high cost of cogeneration equipment, the equipment must run for most of the year in order to pay for
itself. Finally, many electric utilities pay less for power generated by their customers
than the charge for power sold to customers. Thus, cogeneration is typically most costeffective in applications where:
•
•
•
•
the demand for both heat and electricity is substantial and nearly continuous
all heat and electricity generated by the system can be used on site
the cost of electricity is relatively high
the cost of fuel is relatively low
Sizing Combined Heat and Power Systems
In most cases, electrical utilities can generate and sell electricity for less than the cost of
purchasing and operating combined heat and power equipment. In these cases,
combined heat and power equipment only generates net savings when both electricity
and thermal output can be used. Further, combined heat and power equipment only
generates the savings needed to pay for the investment when it is operating. Thus,
except in special cases, combined heat and power systems should be sized based on a
plant’s minimum electricity and heating requirements, so that the cogeneration system
can run continuously at full capacity and all electrical and thermal energy can be
utilized.
Example
A combined heat and power system has an electrical efficiency of 40%, and 60% of the
waste heat can be used in a manufacturing process. Given the load profiles shown
below for a manufacturing plant, determine the maximum size of the system so that all
electrical and thermal power can be used.
3
The combined heat and power system can be modeled as shown below, where Qf is fuel
input, Qw is waste heat, E is electrical output and Qu is useful heat. The electrical
efficiency is Eff,e = E / Qf and the heat exchanger efficiency is Eff,hx = Qu / Qw1.
4
Qw2
Qu
Eff,hx
Qw1
Qf
E
Eff,e
Combining energy balances and efficiency relations gives the following equation for Qu
as in terms of E, Eff,e and Eff,hx:
Qu = E Eff,hx (1 – Eff,e) / Eff,e
(1)
For the load profiles shown above, the maximum electrical power that can be utilized
continuously is 50o kW. From Equation 1, the useable thermal power, Qu, from a CHP
system with 50o kW electrical power is:
Known E, Solve for Qu
INPUTS
Eff,e
Eff,hx
E (kW)
0.4
0.6
500
CALCULATIONS
Qu (kW) = E Eff,hx (1-Eff,e) /Eff,e
450
Based on this calculation, at 500 kW of electrical power output, the useful thermal
output would be 450 kW. However on a continuous basis, the plant can only use 30 kW
of thermal power. Thus, useful power of 300 kW becomes the constraint. Solving
Equation 1 for the electrical power output, E, at the useful thermal power constraint of
300 kW gives:
5
Known Qu, Solve for E
INPUTS
Eff,e
Eff,hx
Qu (kW)
0.4
0.6
300
CALCULATIONS
E (kW) = Qu / [ Eff,hx (1-Eff,e)/Effe ]
333
Thus, the largest CHP system in which all the electrical and thermal power can be used is
Emax = 333 kW and Qu,max = 300 kW.
Types of CHP Systems
Most CHP technologies fall into three categories:



Steam to Power
Power to Heat
Heat to Power
Steam-to-Power Systems
Many manufacturing processes require large quantities of steam. In these cases, it may
be economical to generate power from the steam before using it in the process. A
typical steam-to-power cogeneration system employs a boiler to generate high-pressure
steam. The high-pressure steam is fed into a turbine to generate shaft work. The shaft
work drives an electrical generator and creates electrical power. Some (or all) of the
steam is extracted from the turbine at a pressure high enough to be used in the plant.
The rest of the steam may be extracted at less than atmospheric pressure, condensed
and returned to the boiler.
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Qcoal
Make-up water
Boiler
Shaft
work
out
Turbine
Pump
Condensor
Electrical
Generator
Electricity
Low-pressure steam
~ 50 - 150 psig steam to plant
Qcondensor
Steam turbine performance is typically characterized in terms of isentropic efficiency.
The isentropic efficiency of a turbine is the ratio of actual energy output to maximum
energy output from an isentropic turbine. Assume steam enters a turbine at statepoint
(5) and exits the turbine at statepoint (6). In an ideal isentropic turbine with no internal
friction or turbulence, the steam would exit the turbine at statepoint (6s), which has the
same entropy as statepoint (5). Thus the isentropic efficiency, Effturbine, can be defined
as:
Effturbine = We,actual / We,max = ms (h5 – h6) / ms (h5 – h6s) = (h5 – h6) / (h5 – h6s)
Thus, the enthalpy of steam exiting a turbine (6), can be determined from the entering
enthalpy h5 and isentropic efficiency as:
h6 = h5 – (h5 – h6s) x Effturbine
According a back-pressure turbine manufacturers, packaged turbine/generator sets
include the turbine, generator, skid, speed-reduction gears, controls and switch gear.
The combined efficiency of the speed-reduction gear and the electrical generator is
about 95%, and the cost of a turbine generator set is about $350 per kW.
Example
A large coal boiler produces 265,000 pounds per hour of saturated steam at 180 psig,
for a large plant that operates 8,000 hours per year. The boiler is 80% efficient. The
boiler feed water is a saturated liquid at 140 F. The boiler is capable of producing steam
at up to 700 psig. This capability could be utilized to generate electricity while still
meeting the process heating requirements of the plant.
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Neglecting flash losses from the deaerator tank, calculate the annual additional cost of
coal ($/year), annual electricity cost savings ($/year) and net annual savings ($/year)
from producing 600-psig steam in the boiler and generating electricity from a backpressure turbine / generator set. The isentropic efficiency of a back-pressure turbine is
65%. The combined efficiency of the speed-reduction gear and the electrical generator
is 95%. The cost of coal is $4.00 /mmBtu and the cost of electricity is $0.06 /kWh.
Current System
2
Qexhaust
Qfuel
Deaerator Tank
Process
Heat
Exchanger
Boiler
Qprocess
3
1
Steam Trap
4
Feed Water Pump
Proposed System
Qgen
We
5
Turbine
Ws
Qexhaust
Electrical
Generator
6
Qfuel
Deaerator Tank
Boiler
Process
Heat
Exchanger
Qprocess
3
1
Steam Trap
4
Feed Water Pump
8
Current System
The boiler feedwater (1) is a saturated liquid at 140 F. The steam leaving the boiler (2) is
a saturated vapor at 195 psia. The boiler is 80% efficient. Thus, current fuel use is:
Qf1 = ms x (h2 - h1) / Effboiler
Qf1 = 265,000 lb/hr x (1,199 – 108) (Btu/lb) / 0.80 = 361 mmBtu/hr
Assuming that heat to process is delivered through heat exchangers and the condensate
leaves the heat exchanger (3) as a saturated liquid at 195 psia, the heat delivered to the
process is:
Qp= ms x (h2 – h3)
Qp = 265,000 lb/hr x (1,199 – 353) Btu/lb = 224 mmBtu/hr
Proposed System
In the proposed system, steam would enter the turbine (5) as a saturated vapor at 615
psia. The isentropic efficiency of the turbine is 65%. Thus, the enthalpy of steam exiting
the turbine (6), would be:
h6 = h5 – (h5 – h6s) x Effturbine
h6 =1,203 Btu/lb – (1,203 Btu/lb – 1,111 Btu/lb) x 65% = 1,143 Btu/lb
The quality of the steam exiting the turbine (6) at 105 psia and this enthalpy would be
about 93%. After passing through heat exchangers to deliver heat to process,
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condensate leaves the process heat exchangers (7) as a saturated liquid at 195 psia. The
quantity of steam required to produce the same quantity of process heat with steam
exiting the turbine at this condition would be about:
ms2 = Qp / (h6 - h7)
ms2 = 224 mmBtu/hr / (1,143 Btu/lb – 353 Btu/lb) = 284,000 lb/hr
The steam exiting the boiler (5) would be a saturated vapor at 615 psia with an enthalpy
of about 1,203 Btu/lb. The quantity of fuel required to produce this quantity of steam
would be about:
Qf2 = ms2 x (h5 - h1) / Effboiler
Qf2 = 284,000 lb/hr x (1,203 – 108) (Btu/lb) / 0.80 = 389 mmBtu/hr
Net Savings
The additional fuel for the boiler in the proposed system would be:
dQf = Qf2 – Qf1
dQf = 389 mmBtu/hr – 361 mmBtu/hr = 28 mmBtu/hr
The annual addition fuel and fuel cost would be about:
dQf = 28 mmBtu/hour x 8,000 hour/year = 224,000 mmBtu/year
dCf = 224,000 mmBtu/year x $4.00 /mmBtu = $896,000 /year
The combined efficiency of the speed-reduction gear and the electrical generator is
95%. Thus, the electrical output produced by the turbine / generator would be:
We = ms2 x (h5 - h6) x Effgen
We = 284,000 lb/hr x (1,203 Btu/lb -1,143 Btu/lb) x 95% / 3,413 Btu/kWh = 4,743 kW
The annual electricity and electricity cost savings would be about:
We = 4,743 kW x 8,000 hour/year = 37,944,000 kWh/year
Ce = 37,944,000 kWh/year x $0.06 /kWh = $2,276,640 /year
The net annual cost savings would be about:
$2,276,640 /year - $896,000 /year = $1,380,640 /year
Simple Payback
Turbine/generator sets typically include the turbine, generator, skid, speed-reduction
gears, controls and switch gear, and cost about $350 per kW. If shipping and installation
doubles the purchase cost, the total installed cost of a 5,000 kW system would be about:
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[(5,000 kW x $350 /kW)] x 2 = $3,500,000
The simple payback would be:
$3,500,000 / $1,380,640 /year x 12 months/year = 30 months
The overall thermal efficiency of the system, Effthermal, is:
We = 4,743 kW x 3,413 Btu/kWh / 1,000,000 Btu/mmBtu = 16 mmBtu/hr
Effthermal = (We + Qp) / Qf2
Effthermal = [16 mmBtu/yr + 224 mmBtu/hr] / 389 mmBtu/hr = 62%
Power to Thermal Energy Systems
Power-to-thermal energy systems typically employ a gas turbine, diesel engine or fuel
cell to generate electrical power. The waste heat from the power generation device is
then directed to a boiler or heat exchanger which converts the waste heat into useful
heat for the plant.
Utilizing Heat from Exhaust Gasses
A common method to utilize the heat from the hot exhaust gasses from a gas turbine or
diesel engine is to generate steam or hot water. Standard steam boilers are designed to
use high-temperature (~3,000 F) combustion gasses at relatively low flow rates. The
exhaust gas from a turbine is typically at much lower temperatures (600 – 1,000 F) and
higher flow rates. Thus, it is advisable to purchase a specially designed boiler for most
cogeneration systems. “Waste heat” boilers usually typically have the capacity for
supplemental heating to provide steam when the turbine is not operating and to boost
the temperature of the exhaust gasses. The cost of cogeneration boilers is typically
about $15 per pound of steam per hour.
The quantity of useful heat reclaimed by a waste heat boiler, Qu, can be calculated as:
Qu = mex x cpex x (Tex from turbine/engine – Tex from boiler)
where the exhaust mass flow rate, mex, and temperature, Tex from turbine, are typically
given as turbine performance specifications. The specific heat of natural gas exhaust,
cpex, is about 0.26 Btu/lb-F.
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Gas Turbines
A typical gas turbine cogeneration system is shown below. Ambient air is compressed
to a high pressure, then heated to further raise the pressure. The hot, high pressure air
is then expanded through a turbine and discharged to ambient at between 600 F and
1,000 F. The shaft work generated by the turbine is sufficient to power the compressor
and turn an electrical generator. The generator set typically includes a speed reduction
gear and power conditioning equipment so that the electrical power output is threephase AC power that can be fed directly into the plant.
Qnatural gas
Combustor
Shaft
work
out
Shaft work
to compressor
Compressor
Electrical
Generator
Turbine
Electricity
Exhaust at ~375 F
Turbine exhaust at
650 – 1,000 F
Ambient air
Steam
Boiler
Significantly higher efficiency can be achieved by adding a recuperator to pre-heat air
entering the combustor from the hot exhaust gasses. A schematic of a turbine
generator set with recuperator is shown below.
Exhaust Out
Air In
Heat Exchanger
Natural Gas In
Burner
Turbine
Shaft
Compressor
Shaft
Electricity Out
Electrical Generator
In CHP systems, it is sometimes useful to designate electrical power output from the
generator with unit kWe and shaft power output from the turbine with unit kW.
Nominal performance data from turbine generator sets are shown in the table below.
Nominal performance data were taken at standard conditions of sea level air at 59 F.
Turbine efficiency and power output degrade at higher elevations and as the inlet air
temperature increases. In addition, turbine efficiency also declines when the turbines
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are operated at less than rated output. Gas turbine lifetime is indefinite with proper
maintenance because maintenance includes replacement parts. Typical maintenance
costs are about $0.004 /kWh.
Saturn 20
Centaur 40
Mercury 50
Electrical Output
1,210 kWe
3,515 kWe
4,600 kWe
Heat Rate
14,025 Btu/kWhe 12,245 Btu/kWhe 8,863 Btu/kWhe
Exhaust Flow
51,890 lb/hr
150,715 lb/hr
140,400 lb/hr
Exhaust Temp
940 F
830 F
710 F
Source: www.mysolar.cat.com
The electrical efficiency of the turbines is given by the reciprocal of the heat rate. The
Mercury 50 turbine employs a recuperator to preheat air to the burner and has a
significantly lower heat rate. The electrical efficiency of the Saturn 20 and Mercury 50
turbine generator sets are about:
EffSaturn20 = 1 / Heat rate = (1 / 14,025 Btu/kWh) x 3,412 Btu/kWh = 24%
EffMercury50 = 1 / Heat rate = (1 / 8,863 Btu/kWh) x 3,412 Btu/kWh = 38%
Εxample
A turbine generator set has the following performance data.
Electrical Output:
Heat Rate:
Exhaust Flow:
Exhaust Temp:
4,162 kWe
8,544 Btu/kWhe
136,000 lb/hr
663 F
Calculate the thermal output at 100% load, the energy recoverable by directing exhaust
gasses through a heat recovery boiler, and the overall thermal efficiency. The exhaust
gasses leave the boiler at 375 F.
The fuel energy input, Qf, at 100% load is:
Qf = 4,162 kW x 8,544 Btu/kWh = 35.56 mmBtu/hr
The electrical energy output, We, at 100% load is:
We = 4,162 kW x 3,412 Btu/kWh x 1 mmBtu/ 106 Btu = 14.20 mmBtu/hr
The specific heat of the exhaust air is 0.26 Btu/lb-F. The temperature of the exhaust
leaving a cogeneration boiler is 375 F. The useful energy to the steam, Qu, is:
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Qu = m cp (Texhaust,turbine – Texhaust,boiler)
Qu = 136,000 lb/hr x 0.26 Btu/lb-F x (663 – 375) F = 10.18 mmBtu/hr
The overall thermal efficiency of the system, Effthermal, is:
Effthermal = (Welec + Qu) / Qf
Effthermal = [14.20 mmBtu/hr + 10.18 mmBtu/hr] / 35.56 mmBtu/hr = 69%
Micro Turbines
Micro turbines with integrated electrical generators with electrical power output
ranging from 30 kW to 200 kW are also available. Multiple micro turbines can be staged
in applications with variable loads to reduce part load efficiency losses. A 65-kW
microturbine may use 0.84 mmBtu per hour of natural gas at full load and discharge 600
scfm of exhaust air at 800 F.
30 kW micro turbine
www.capstonemicroturbine.com
Reciprocating Engines
Reciprocating engines fueled by natural gas or diesel fuel can be coupled to electrical
generators as shown below. These engines range in size from about 10 kWe to 20
MWe.
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Diesel engine with electrical generator set. Source: www.cat.com.
Heat is rejected from reciprocating engines through several pathways. An approximate
breakdown of heat loss from a diesel powered electrical generation set is shown below.
In general, heat loss from the hot surfaces of the engine and generator is difficult to
reclaim. Heat loss in the exhaust and coolant is easier to reclaim, however the
maximum temperature of the coolant is generally so low (~ 120 F) that it limits heat
reclaim possibilities.
Heat rejection from a diesel generator set at full load (from www.cat.com)
Heat rejection to coolant (total)
Heat rejection to exhaust (total)
Heat rejection to after cooler
Heat rejection to atmosphere from engine
Heat rejection to atmosphere from generator
21%
52%
11%
12%
3%
Example:
An 830 kWe diesel generator set consumes 57.6 gal/hr of fuel, produces 6,397cfm of
exhaust at 847 F. Calculate the rate of useful heat that can be reclaimed from the
exhaust if the exhaust leaves a heat reclaim heat exchanger at 400 F. Calculate the total
thermal efficiency of the system. The product of the density and specific heat of
exhaust gasses is about 0.018 Btu/lb-F and the energy content (HHV) of diesel fuel is
130,500 Btu/gal.
The rate of useful heat that can be reclaimed from the exhaust, Qu, if the exhaust is:
INPUTS
V (cfm)
pcp (Btu/ft3-F)
Tex1
Tex2
CALCULATIONS
Qu (Btu/hr) = V pcp (Tex1 - Tex2) 60
Qu (kW) = Qu (Btu/hr) / 3,412
6397
0.018
847
400
3,088,216
905
The fuel energy input, Qf, and total thermal efficiency Eff,thermal, are:
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INPUTS
Vf (gal/hr)
HHV (Btu/gal)
We (kW)
CALCULATIONS
Qf (Btu/hr) = Vf HHV
Qf (kW) = Qf (Btu/hr) / 3,412
Eff,thermal = (We + Qu) / Qf
57.6
130,500
830
7,516,800
2,203
0.79
Gas-Fired Reciprocating Engines:
Gas-fired reciprocating internal combustion engines offer many advantages over other
technologies for small scale CHP such as low capital cost, easy start-up, high reliability,
good load-following characteristics, less expensive backup power, and the ability to
generate onsite power during peak demand periods. Commercially available natural gas
reciprocating engines produce from 0.5 kW to 10 MW of electrical power.
Natural Gas Fired Reciprocating Engine. Source: http://www.allforpower.com
Heat can be recovered from the exhaust gas, intercooling stages, lube oil and jacket
water. Heat loss from the hot surfaces of the engine and generator is difficult to
reclaim. Specifications of a typical natural gas-fired reciprocating engine are shown in
the following table. At full load, 41.5% of fuel energy is converted into electrical power
and 44.8% into useful recoverable heat. Thus, the total efficiency is 86.3%.
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Technical Data
Full Load (100%) Medium Load (75%) Low Load (50%)
14.73
11.38
8.04
1.70
1.31
0.93
2.51
1.88
1.26
1.79
1.33
0.88
Recoverable Thermal Output
Intercooler first stage (kBtu/hr)
1.34
0.75
0.22
Lube oil (with gearbox) (kBtu/hr)
0.71
0.63
0.55
Jacket water (kBtu/hr)
1.06
0.90
0.78
Exhaust gas cooled to 248 F (kBtu/hr)
3.49
2.96
2.28
Total Recoverable Thermal Output (kBtu/hr)
6.60
5.25
3.83
Efficiencies
Electrical Efficiency
41.5%
39.8%
37.3%
Thermal Efficiency
44.8%
46.1%
47.7%
Total Efficiency
86.3%
85.9%
85.0%
Hot Water Circuit
Forward Temperature (F)
194
170
167
Return Temperature (F)
158
158
158
Hot water flow rate (gpm)
367
367
367
Parameter (units)
Energy Input (kBtu/hr)
Gas Volume (scfhr)
Mechanical Output (bhp)
Electrical Output (kW)
Mass flow rates on combustion air and exhaust gases and temperature of the exhaust
are shown in the following table. This information can be used to determine the amount
of energy that can be recovered from the exhaust gas stream.
Combustion Air and Exhaust Gas Data
Parameter
Value
Units
Combustion air mass flow rate
22.6 lbs/hr
Exhaust Gas temperature at full load
804 F
Exhaust Gas Mass flow rate (wet)
23.3 lbs/hr
Exhaust Gas Mass flow rate (dry)
21.8 lbs/hr
Exhaust gas volume, wet
310.4 scfhr
Exhaust gas volume, dry
280.4 scfhr
Example:
A natural gas reciprocating engine consumes 1.7 scfm/hr of natural gas at full load, and
produces 23 lbs/hr of exhaust at 804 F. A facility receives city water at an average
annual temperature of 60 F and is considering installing an 80% effective counter flow
heat exchanger to produce hot water. Assuming the mass flow rate of water is the same
as the exhaust gas, what would be the temperature of exiting hot water and the amount
of energy recovered through the heat exchange process? The specific heats of the
exhaust gas and water are about 0.26 Btu/lb-F and 1.0 Btu/lb-F respectively.
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HX WITH KNOWN EFFECTIVENESS
INPUTS
Term
Value
Incoming Hot Fluid Temperature, Th1
Hot Fluid Mass Flow Rate, mh
Hot Fluid Specific Heat, cph
Incoming Cold Fluid Temperature, Tc1
Cold Fluid Mass Flow Rate, mc
Cold Fluid Specific Heat, cpc
Heat Exchanger Effectiveness, e
Units
804 F
0.4 lb/min
0.26 Btu/lb-F
60 F
0.4 lb/min
1 Btu/lb-F
80%
CALCULATIONS
Term
Hot Fluid Heat Capacity, Ch = mh * cph
Cold Fluid Heat Capacity, Cc = mc * cph
Cmin = min(Ch,Cc)
Cmax = max(Ch,Cc)
Heat Transfer, Q = e*Cmin*(Th1-Tc1)
Exiting Hot Fluid Temperature, Th2 = Th1-Q/Ch
Exiting Cold Fluid Temperature, Tc2 = Tc1 + Q/Cc
Heat Transfer, Q
Value
Units
0.1 Btu/min-F
0.4 Btu/min-F
0.1 Btu/min-F
0.4 Btu/min-F
59 Btu/min
209 F
208 F
3,559 Btu/hr
Incoming and outgoing temperatures of both exhaust (Th1, Th2) and water (Tc1, Tc2) are shown
in the following diagram.
804 F
Th1
Tc2
208 F
Tc1
60 F
Qact
209 F
Th2
Heat to Power
Rankine power cycles use heat to make electrical power. In traditional Rankine systems,
the working fluid is water. At atmospheric pressure water boils at 212 F, and the boiling
temperature increases at the higher pressures that generate more work; thus steampowered Rankine cycles require relatively high temperature heat to be effective.
In contrast, refrigerants boil at much lower temperatures than water. Thus, Rankine
cycles using refrigerant as the working fluid are able to generate power using much
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lower temperature heat. These systems are sometimes called “organic Rankine
systems” with reference to the organic nature of the refrigerants.
Many industrial plants generate large quantities of relatively low-temperature waste
heat. In these cases, low-temperature heat-to-power systems can use the heat to
generate electrical power.
Low-temperature organic Ranking cycle. Source: www.transpacenergy.com
An organic Rankine system that uses heat reclaimed from exhaust gasses is shown in the
figure below. Heat is reclaimed from hot exhaust by an air-to-liquid heat exchanger.
The hot liquid leaves the heat exchanger at 265 C and vaporizes refrigerant in the
vaporizer. The vapor is expanded by a turbine, producing shaft work that drives an
electrical generator. The warm low-pressure vapor leaving the turbine delivers some
heat to the low-temperature refrigerant before being condensed in an air-cooled
condenser. A pump increases the pressure of the low temperature fluid, before it picks
up heat from the hot vapor leaving the turbine. The fluid then returns to the vaporizer
and the cycle is repeated.
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Source: Muller, M. “Organic Rankine Cycles”, IAC Director’s Meeting, San Francisco, CA,
2010.
Another system is shown below. This system can use heat at a temperature as low as
121 C, and an evaporative cooler instead of an air cooler to condense the working fluid.
The system use R245fa as the working fluid.
20
Source: Muller, M. “Organic Rankine Cycles”, IAC Director’s Meeting, San Francisco, CA,
2010.
Self-contained modular organic Rankine power systems are now on the market. System
sizes ranges from about 10 kW to about 300 kW. Typical systems can use hot water in
the 190 F range or exhaust gasses in the 400 F range. Typical installed system cost is
about $2,500 per kWe.
Example
A modular organic Rankine system costs $2,500 /kW and generates 50 kW of electrical
power 8,000 hours per year. If the electricity generated by the system displaces
electricity purchased from a utility at $0.10 /kW, calculate the simple payback of the
system.
Cost savings = 50 kWe x 8,000 hr/yr x $.10 /kWh = $40,000 /year
Implementation cost = 50 kW x $2,500 /kW = $125,000
Simple payback = $125,000 / $40,000 /year = 3.125 years
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